$f(t) = -4t^{2}-4t-3(h(t))$ $h(x) = -x^{2}-5(g(x))$ $g(t) = -4t+7$ $ h(g(-6)) = {?} $
Explanation: First, let's solve for the value of the inner function, $g(-6)$ . Then we'll know what to plug into the outer function. $g(-6) = (-4)(-6)+7$ $g(-6) = 31$ Now we know that $g(-6) = 31$ . Let's solve for $h(g(-6))$ , which is $h(31)$ $h(31) = -31^{2}-5(g(31))$ To solve for the value of $h$ , we need to solve for the value of $g(31)$ $g(31) = (-4)(31)+7$ $g(31) = -117$ That means $h(31) = -31^{2}+(-5)(-117)$ $h(31) = -376$